WATT GOVERNOR

Watt Governor

The simplest form of a centrifugal governor is a Watt governor, as shown in Fig.  It is basically a conical pendulum with links attached to a sleeve of negligible mass. The arms of the governor may be connected to the spindle in the following three ways :

1.      The pivot P, may be on the spindle axis as shown in Fig.

2.       The pivot P, may be offset from the spindle axis and the arms when produced intersect at O, as shown in Fig. 18.2 (b).

3.      The pivot P, may be offset, but the arms cross the axis at O, as shown in Fig. . 

Let

m  = Mass of the ball in kg,

w

= Weight of the ball in newtons = m.g,

T

= Tension in the arm in newtons,

ω

= Angular velocity of the arm and

ball about the spindle axis in

rad/s,

r

= Radius of the path of rotation of the ball i.e. horizontal distance

from the centre of the ball to the

spindle axis in metres,

F

C

= Centrifugal force acting on the ball in newtons  = m . ω 2.r, and

h

= Height of the governor in metres.

It is assumed that the weight of the arms, links and the sleeve are negligible as compared to the weight of the balls. Now, the ball is in equilibrium under the action of

1. the centrifugal force (F_C) acting on the ball, 2.  the tension (T) in the arm, and 3. the weight

(w) of the ball.
Taking moments about point O, we have
		FC × h	= w × r  =  m.g.r
or	m. ω 2.r.h		= m .g.ror				h = g / ω 2		. . . (i)
When g is expressed in m/s2 and ω in rad/s, then h is in metres. If N is the speed in r.p.m., then
		ω	= 2 π N/60
			9.81			895
∴		h =			=			metres	. . . (∵ g = 9.81 m/s2) . . . (ii)
				(2 π N / 60)2		N 2

Note : We see from the above expression that the height of a governor h, is inversely proportional to N ². Therefore at high speeds, the value of h is small. At such speeds, the change in the value of h corresponding to a small change in speed is insufficient to enable a governor of this type to operate the mechanism to give the necessary change in the fuel supply. This governor may only work satisfactorily at relatively low speeds i.e. from 60 to 80 r.p.m.

Calculate the vertical height of a Watt governor when it rotates at 60 r.p.m. Also find the change in vertical height when its speed increases to 61 r.p.m.

Solution. Given : N₁ = 60 r.p.m. ; N₂ = 61 r.p.m.

Initial height

We know that initial height,

h	=	895	=	895	= 0.248 m
1		( N ) 2	(60)2
	1

Change in vertical height

We know that final height,

h	=	895	=	895	= 0.24 m
2		( N 2 ) 2	(61)2

∴ Change in vertical height

h₁ – h₂ =  0.248 – 0.24  =  0.008 m  =  8  mm Ans.